Insert $(9^*)$ in $(11)$ and solve for $C$ as a function of $K$, using $(8)$ for $r$. Substitute back in $(9^*)$ and solve for $K$ as a function of $L$. Insert into $(12)$ and you will get $w$ as a function of parameters. Etc.

$\begingroup$Thanks @Alecos. I don’t understand what you do with equation (10). At the end, $C^{\sigma}$ is equal to what?$\endgroup$
– TeconJan 26 '19 at 20:41

$\begingroup$@Tecon I start with the per capita expression and then write $c=C/N, l = L/N$, which gives a denominator of $N^{\sigma + \gamma}$ that I then move to the other side of the equation.$\endgroup$
– Alecos PapadopoulosJan 26 '19 at 22:46

$\begingroup$Sorry for my previous stupid question, I was reading from the phone and I couldn't see the end of the equation. I followed your steps. Given that $N=1$, equation (10*) is equal to equation (10). I substitute (9*) in (11) and i get: $C=\frac{r}{\alpha}K$. I then substitute $C$ in (9*) and obtain: $K=(\frac{\alpha}{r})^{(\frac{1}{1-\alpha})}L$. I plug this into (12) and obtain $r = \alpha (1-\alpha)^{\frac{1-\alpha}{\alpha}}w^{\frac{\alpha-1}{\alpha}}$.$\endgroup$
– TeconJan 28 '19 at 11:02

$\begingroup$This last expression is identical to the one obtained by "solving for the ratio", as I show here. How do you interpret this?$\endgroup$
– TeconJan 28 '19 at 11:03